Boosting power supply available current

I just finished constructing an adjustable Power Supply Unit (PSU) using the common and highly-available LM317 regulator. I intended to use this supply to power light loads, such as stomp boxes and small audio effect units. Of course, a moment had come, when I needed to power a heavier load; a small power amplifier I designed. The PSU I created could not drive that demanding load  and I had to reconstruct the circuit of the supply.

This time I added a current booster circuit using the TIP32C transistor with the LM317 regulator to be able to drive heavy loads. The transformer is not shown in the schematics, only its output that is 12VAC.

In Fig.1 the original power supply circuit is showed (without current booster).

Fig.1 LM317 adjustable power supply with 100kΩ load

Using a load of 100kΩ all things run nice and we get the following output (Fig.2)

Fig.2 DC output voltage with 100kΩ load

Now, if we decrease the value of the load resistor - making a heavier and current demanding load - we get the following graph (Fig.3).

Fig.3 DC output voltage with 4Ω load

We see that the output voltage sagged (dropped) to about 8VDC, indicating that the power supply cannot supply this heavy load the current it requires.

The current boosted Power Supply I designed is seen in Fig.4.

Fig.4 Adjustable Power Supply with current booster using TIP32C

The concept is that when the LM317 comes to its limits, the TIP32C power transistor is turned on to supply extra current to the load in parallel with the LM317. Of course I used 4 TIP32C in parallel to increase the current boost available. The power transistors turn on when there is enough voltage drop across resistor R1 to overcome their base emitter voltage Vbe. (Such voltage drop in the R1 means that the LM317 has come to its limits and the power supply unit starts sagging).

In Fig.5 a plot is seen for a heavy load of 4Ω.

Fig.5 DC output voltage for current boosted power supply

We seen now that the power supply does not sag. Instead to delivers the required current to the 4Ω load, maintaining the required output voltage at ~12VDC and delivering ~12/VDC/4Ω = ~3Α current.

The ripple voltage of the output is at about ~1V which is acceptable enough, because it is below the standard 10% limit and will not become a noise figure for amplifiers having high enough Power Supply Rejection Ratio PSRR (99% of the available solid state amplifiers). Of course the ripple can be further reduced and the simplest method to do that is to enlarge the reservoir capacitor C1. Increasing it though, the inrush current sucked by the MAINS will be increased, unless a soft-start circuit or a suitable trasformer (saturation limit, windings, etc) is placed. In Fig.6 a plot for a larger C1 is shown. Increasing C1 to 22000μF decreases ripple current to about ~0.5V.

Fig.6 Current boosted power supply with larger reservoir capacitor

Fig.7 Input current requirement for current boosted power supply

The power dissipated in the load is about 12VDC*3A = 36W. The input current for the power supply required to drive such a load is sen in Fig.7 (with C1=10000μF). The rms current is at about 5A, with peaks at about 12.5A. That means if you have a fast MAINS fuse it will turn off. 12VAC*5A = 60W power is required from the transformer before the circuit shown here.  Of course the transformer will worsen the things a little bit. 60W for 36W is the prise we pay for a very simple linear power supply circuit. There are many methods for increasing power supply efficiency, but this is not the purpose of this power supply, since it is a test equipment.